Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 35

The decomposition of \(\mathrm{HI}(\mathrm{g})\) at \(700 \mathrm{K}\) is followed for \(400 \mathrm{s},\) yielding the following data: at \(t=0,[\mathrm{HI}]=\) \(1.00 \mathrm{M} ;\) at \(t=100 \mathrm{s},[\mathrm{HI}]=0.90 \mathrm{M} ;\) at \(t=200 \mathrm{s}, [\mathrm{HI}]=0.81 \mathrm{M} ; t=300 \mathrm{s},[\mathrm{HI}]=0.74 \mathrm{M} ;\) at \(t=400 \mathrm{s}, [\mathrm{HI}]=0.68 \mathrm{M} .\) What are the reaction order and the rate constant for the reaction: $$\mathrm{HI}(\mathrm{g}) \longrightarrow \frac{1}{2} \mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{I}_{2}(\mathrm{g}) ?$$ Write the rate law for the reaction at 700 K.

Short Answer

Expert verified
The reaction order is 1 (first order). The rate law for the reaction at 700 K is \[Rate = k[HI]\].

Step by step solution

01

Determine the reaction order

To deduce the reaction order, we observe the behavior of the concentration of \(HI\) over time. We can see that the concentration of \(HI\) is decreasing by a factor of 0.9 for every 100 s time interval (from 1.00 M to 0.90 M, then 0.90 M to 0.81 M and so on), suggesting a first-order reaction.
02

Calculate the rate constant using the first-order reaction formula

For a first-order reaction, the rate constant \(k\) can be calculated using the formula: \[k = \frac{1}{t} ln\frac{[HI]_0} {[HI]_t}\] where \([HI]_0\) is the initial concentration of \(HI\), \([HI]_t\) is the concentration of \(HI\) at a certain time \(t\). Given that \([HI]_0\) = 1.00 M, \([HI]_t\) = 0.90 M, and \(t\) = 100 s, we can substitute these values into the formula: \[k = \frac{1}{100 s} ln\frac{1.00 M}{0.90 M}\]
03

Write the rate law of the reaction

Once we have determined the order of the reaction and the rate constant, we can write the rate law of the reaction. For a first-order reaction involving a reactant \(HI\), the rate law is: \[Rate = k[HI]\] where \(k\) is the rate constant we calculated and \([HI]\) is the concentration of \(HI\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

For the reaction \(A+B \longrightarrow C+D\), the following initial rates of reaction were found. What is the rate law for this reaction? $$\begin{array}{llll} \hline & & & \text { Initial Rate, } \\ \text { Expt } & \text { [A], M } & \text { [B], M } & \text { M min }^{-1} \\\ \hline 1 & 0.50 & 1.50 & 4.2 \times 10^{-3} \\ 2 & 1.50 & 1.50 & 1.3 \times 10^{-2} \\ 3 & 3.00 & 3.00 & 5.2 \times 10^{-2} \\ \hline \end{array}$$

The reaction \(A+B \longrightarrow\) products is first order in \(A\) first order in \(\mathrm{B},\) and second order overall. Consider that the starting concentrations of the reactants are \([\mathrm{A}]_{0}\) and [ \(\mathrm{B}]_{0},\) and that \(x\) represents the decrease in these concentrations at the time \(t .\) That is, \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-x\) and \([\mathrm{B}]_{t}=[\mathrm{B}]_{0}-x .\) Show that the integrated rate law for this reaction can be expressed as shown below. $$\ln \frac{[\mathrm{A}]_{0} \times[\mathrm{B}]_{t}}{[\mathrm{B}]_{0} \times[\mathrm{A}]_{t}}=\left([\mathrm{B}]_{0}-[\mathrm{A}]_{0}\right) \times k t$$

For the first-order reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ \(t_{1 / 2}=22.5 \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) and \(1.5 \mathrm{h}\) at \(40^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) If the Arrhenius constant \(A=2.05 \times 10^{13} \mathrm{s}^{-1}\) determine the value of \(k\) at \(30^{\circ} \mathrm{C}\).

Derive a plausible mechanism for the following reaction in aqueous solution, \(\mathrm{Hg}_{2}^{2+}+\mathrm{Tl}^{3+} \longrightarrow 2 \mathrm{Hg}^{2+}+\mathrm{Tl}^{+}\) for which the observed rate law is: rate \(=k\left[\mathrm{Hg}_{2}^{2+1}\right]\) \(\left.[\mathrm{T}]^{3+}\right] /\left[\mathrm{Hg}^{2+}\right].\)

In three different experiments, the following results were obtained for the reaction \(A \longrightarrow\) products: \([\mathrm{A}]_{0}=1.00 \mathrm{M}, t_{1 / 2}=50 \mathrm{min} ;[\mathrm{A}]_{0}=200 \mathrm{M}, t_{1 / 2}=\) \(25 \min ;[\mathrm{A}]_{0}=0.50 \mathrm{M}, t_{1 / 2}=100 \mathrm{min} .\) Write the rate equation for this reaction, and indicate the value of \(k.\)
See all solutions

Recommended explanations on Chemistry Textbooks

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Chemical Analysis

Read Explanation

Physical Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free